# Projection speed mechanics

1. Aug 13, 2009

### kfox1984

i have to answer this question for an assignment that I need to do for mechanics. I am really really stuck - would somebody please mind helping....

A ball thrown at an angle α to the horizontal just clears a wall. The horizontal and vertical distances to the top of the wall are X and Z respectively, with X.tan α>Z. Show that the projection speed is...

X/cos α { g/(2(X.tanα – Z)}^1/2

Where g is acc due to gravity. Determine the position of the highest point reached in terms of X. Z and tan α.

Last edited by a moderator: May 11, 2014
2. Aug 13, 2009

### Staff: Mentor

Hint: Write equations for the vertical and horizontal positions as functions of time.

3. Aug 13, 2009

### kfox1984

Thanks - I have already done that but still nothing....

4. Aug 13, 2009

### Staff: Mentor

Show what you did. Make use of the fact that when x = X, y = Z.

5. Aug 13, 2009

### kfox1984

I have done that -
i'm not sure which equation to use...

I know that tx = ty so i have tried rearranging things and substituting. but i still have v's left in everything....

i dont understand where the tan has come from...

I'm guessing that i have to use v^2 = u^2 + 2as as the final eqn has a 2 in it but i dont understand how the two has got to the bottom of the fraction sign. I also dont understand where the squre root has come from.......

6. Aug 13, 2009

### kfox1984

I dont even know if i'm supposed to be using calculus or not - this is supposed to be an as level mechanics problem

7. Aug 13, 2009

### Staff: Mentor

Again I ask you to write down the expressions for horizontal (x) and vertical (y) positions as a function of time. (They will also involve V and α.) The tan and the 2 will fall into place automatically when you later set up and solve those equations.

(Nothing to do with calculus, just a little algebra.)

8. Aug 13, 2009

### kfox1984

xcos(α) = (U + V)/2 * t , zsin(α) = (u + V)/2 * t

9. Aug 13, 2009

### kfox1984

or x = [(ucos(α) + V/)2] *t and z = [(usin(α) + V)/2] * t

10. Aug 13, 2009

### Staff: Mentor

Those aren't correct. Review the kinematic equations here: https://www.physicsforums.com/showpost.php?p=905663&postcount=2"

(The ones you want relate displacement and time.)

Getting closer, but still not right.

Last edited by a moderator: Apr 24, 2017
11. Aug 13, 2009

### kfox1984

I'm sorry but i still really dont understand what i am supposed to do with these.

12. Aug 13, 2009

### Staff: Mentor

$$x = x_0 + v_0 t + (1/2) a t^2$$

How would you apply it to the horizontal and vertical motion of your projectile?

13. Aug 13, 2009

### kfox1984

ok so s = ut + 1/2at^2.....

x = ucos(α)t , Z = usin(α)t 1/2gt^2 but i'm still unsure about what i am supposed to do next...I already tried rearranging the X one in terms of t then substituting it into the other....which didn't work...if i divide them then i loose the cos(α) using pythagoras was still wrong...?

14. Aug 13, 2009

### kfox1984

sorry thats supposed to be z = u.sin(α) - 1/2gt^2

15. Aug 13, 2009

### kfox1984

x/ucos9α) = t

Z = u.sin(α).(x/cos(α)) - 1/2g[(x/ucos(α))^2]

is this correct?

16. Aug 13, 2009

### Staff: Mentor

Try it again. (It worked for me. ) Take the X equation and solve for t, then substitute that into the Z equation.

Good. But you left out a t in the first term on the right hand side.

17. Aug 13, 2009

### Staff: Mentor

Almost. You dropped a u in the second term. But you are on the right track.

18. Aug 13, 2009

### kfox1984

i get -

Z = usinα.(x/cosα) - 1/2g[x/ucosα]^2 but now i have a cos^2 which i'm not sure how to get rid of??
Thanks

19. Aug 13, 2009

### kfox1984

re arranging i get -

u^2 = xthanα - 1/2.g.(x^2/zcos^2α)

Which is still incorrect, i really dont know where i am going wrong sorry

20. Aug 13, 2009